问下这个SQL怎么下
id name like
1 张三 a,b,c,d,e
2 李四 a,b,c,
.
.
.
我要结果
1 张三 a
2 张三 b
3 张三 c
4 张三 d
5 张三 e
2 李四 a
2 李四 b
2 李四 c
.
.
.
这个用SQL 语句怎么实现.
------解决方案--------------------
参考:
SQL code
/*
标题:简单数据拆分(version 2.0)
作者:爱新觉罗.毓华(十八年风雨,守得冰山雪莲花开)
时间:2010-05-07
地点:重庆航天职业学院
描述:
有表tb, 如下:
id value
----------- -----------
1 aa,bb
2 aaa,bbb,ccc
欲按id,分拆value列, 分拆后结果如下:
id value
----------- --------
1 aa
1 bb
2 aaa
2 bbb
2 ccc
*/
--1. 旧的解决方法(sql server 2000)
create table tb(id int,value varchar(30))
insert into tb values(1,'aa,bb')
insert into tb values(2,'aaa,bbb,ccc')
go
--方法1.使用临时表完成
SELECT TOP 8000 id = IDENTITY(int, 1, 1) INTO # FROM syscolumns a, syscolumns b
SELECT A.id, value = SUBSTRING(A.[value], B.id, CHARINDEX(',', A.[value] + ',', B.id) - B.id)
FROM tb A, # B
WHERE SUBSTRING(',' + A.[value], B.id, 1) = ','
DROP TABLE #
--方法2.如果数据量小,可不使用临时表
select a.id , value = substring(a.value , b.number , charindex(',' , a.value + ',' , b.number) - b.number)
from tb a join master..spt_values b
on b.type='p' and b.number between 1 and len(a.value)
where substring(',' + a.value , b.number , 1) = ','
--2. 新的解决方法(sql server 2005)
create table tb(id int,value varchar(30))
insert into tb values(1,'aa,bb')
insert into tb values(2,'aaa,bbb,ccc')
go
--方法1.使用xml完成
SELECT A.id, B.value FROM
(
SELECT id, [value] = CONVERT(xml,'<root><v>' + REPLACE([value], ',', '</v><v>') + '</v></root>') FROM tb
) A OUTER APPLY
(
SELECT value = N.v.value('.', 'varchar(100)') FROM A.[value].nodes('/root/v') N(v)
) B
--方法2.使用CTE完成
;with tt as
(select id,[value]=cast(left([value],charindex(',',[value]+',')-1) as nvarchar(100)),Split=cast(stuff([value]+',',1,charindex(',',[value]+','),'') as nvarchar(100)) from tb
union all
select id,[value]=cast(left(Split,charindex(',',Split)-1) as nvarchar(100)),Split= cast(stuff(Split,1,charindex(',',Split),'') as nvarchar(100)) from tt where split>''
)
select id,[value] from tt order by id option (MAXRECURSION 0)
DROP TABLE tb
/*
id value
----------- ------------------------------
1 aa
1 bb
2 aaa
2 bbb
2 ccc
(5 行受影响)
*/
------解决方案--------------------SQL code
--有表tb, 如下:
/*
id value
----------- -----------
1 aa,bb
2 aaa,bbb,ccc
*/
--欲按id,分拆value列, 分拆后结果如下:
id value
----------- --------
/*
1 aa
1 bb
2 aaa
2 bbb
2 ccc
*/
SELECT TOP 8000 id = IDENTITY(int, 1, 1) INTO # FROM syscolumns a, syscolumns b
SELECT A.id, SUBSTRING(A.[values], B.id, CHARINDEX(',', A.[values] + ',', B.id) - B.id)
FROM tb A, # B
WHERE SUBSTRING(',' + A.[values], B.id, 1) = ','
------解决方案--------------------
SQL code
create table ht
(id int, name varchar(8), [like] varchar(10))
insert into ht
select 1, '张三', 'a,b,c,d,e' union all
select 2, '李四', 'a,b,c'
select a.id,a.name,
substring(a.[like],b.number,charindex(',',a.[like]+',',b.number)-b.number) 'like'
from ht a
inner join master.dbo.spt_values b
on b.number<=len(a.[like]) and b.type='P'